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a) 2(x-3)-3(x-5)=4(3-x)-18
2x-6-3x-15=12-4x-18
2x-3x+4x=12-18+6+15
3x=15
x=15:3
x=5
Vậy x=5
Bài 1: a) \(-2.\left(2x-8\right)+3.\left(4-2x\right)=\left(-72\right)-5.\left(3x-7\right)\)
\(-4x+16+12-6x=-72-15x+35\)
\(-4x-6x+15x=-72+35-16-12\)
\(5x=-65\)
\(x=-\frac{65}{5}\)
\(x=-13\)
b) \(3.\left|2x^2-7\right|=33\)
\(\left|2x^2-7\right|=\frac{33}{3}=11\)
\(\Rightarrow\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}\Rightarrow\orbr{\begin{cases}2x^2=18\\2x^2=-4\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=9\\x^2=-2\left(vl\right)\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm3\\\end{cases}}}\)
Bài 2:
Ta có: \(2n+1⋮n-3\)
\(2n-6+7⋮n-3\)
\(2\left(n-3\right)+7⋮n-3\)
Vì \(2\left(n-3\right)⋮n-3\)
Để \(2\left(n-3\right)+7⋮n-3\)
Thì \(7⋮n-3\Rightarrow n-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
| n-3 | -1 | 1 | 7 | -7 |
| n | 2 | 4 | 10 | -4 |
Vậy.....
hok tốt!!
b)
\(3\left(2x^2-7\right)=33\)
\(\Leftrightarrow2x^2-7=11\)
\(\Leftrightarrow2x^2=18\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow x=\pm3\)
a) -2(2x - 8) + 3(4 - 2x) = -72 - 5(3x - 7)
=> -4x + 16 + 12 - 6x = -72 - 15x + 35
=> -10x + 28 = -37 - 15x
=> -10x + 15x = -37 - 28
=> 5x = -65
=> x = -65 : 5
=> x = -13
b) 3(2x2 - 7) = 33
=> 2x2 - 7 = 33 : 3
=> 2x2 - 7 = 11
=> 2x2 = 11 + 7
=> 2x2 = 18
=> x2 = 18 : 2
=> x2 = 9
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy ...
a) -2(2x-8)+3(4-2x)=-72-5(3x-7)
<=> -4x+16+12-6x=-72-15x+35
<=> -10x+28=-37-15x
<=> -10x+28+37+15x=0
<=> 5x+65=0
<=> 5x=-65
<=> x=-13
b) 3I2x2-7I=33
<=> I2x2-7I=11
<=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}\Leftrightarrow\orbr{\begin{cases}2x^2=18\\2x^2=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=9\\x^2=\frac{-3}{2}\left(ktm\right)\end{cases}\Leftrightarrow}x=\pm3}\)
\(a,-2.\left(2x-8\right)+3.\left(4-2x\right)=-72-5\left(3x-7\right)\)
\(< =>-4x+16+12-6x=-72-15x+35\)
\(< =>-10x+15x=-72+35-16-12=-65\)
\(< =>5x=-65< =>x=\frac{-65}{5}=-12\)
\(b,3.\left|2x^2-7\right|=33\)
\(< =>\left|2x^2-7\right|=\frac{33}{3}=11\)
\(< =>\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}}\)
\(< =>\orbr{\begin{cases}2x^2=11+7=18\\2x^2=-11+7=-4\end{cases}}\)
\(< =>\orbr{\begin{cases}x^2=9\\x^2=-2\end{cases}< =>\orbr{\begin{cases}x=3or-3\\x=\varnothing\end{cases}}}\)